91Ó°ÊÓ

The Binomial Random Variable\({\rm{X}}\)is defined as

\({\rm{X = }}\)the number of a month\({\rm{n}}\)trials

Where conditions one to four on page\({\rm{117}}\)are satisfied (binomial experiment).

The described random variable has Binomial Distribution with parameters

\({\rm{n = 25}}\)and\({\rm{p = P(S)}}\)

(a)

The second parameter is given –

\({\rm{p = 0}}{\rm{.5}}\)

Which means that –

\({\rm{X}} \sim {\rm{Bin(25,0}}{\rm{.5)}}\)

Cumulative Density Function cdf of binomial random variable\({\rm{X}}\)with parameters\({\rm{n}}\)and\({\rm{p}}\)is –

\({\rm{B(x;n,p) = P(X}} \le {\rm{x) = }}\sum\limits_{{\rm{y = 0}}}^{\rm{x}} {\rm{b}} {\rm{(y;n,p),}}\;\;\;{\rm{x = 0,1, \ldots ,n}}\)

The theorem is –

\({\rm{b(x;n,p) = }}\left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{{\rm{(1 - p)}}}^{{\rm{n - x}}}}}&{{\rm{,x = 0,1,2, \ldots ,n}}}\\{\rm{0}}&{{\rm{, otherwise }}}\end{array}} \right.\)

Now compute the requested probability as –

\(\begin{aligned} P(7 \leqslant X \leqslant 18) &= B(18;25,0.5) - B(6;25,0.5) \\ & = 0.993 - 0.007 \\ &= 0.986 \\ \end{aligned}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.986}}\).

(b)

The second parameter is given –

\({\rm{p = 0}}{\rm{.8}}\)

Which means that –

\({\rm{X}} \sim {\rm{Bin(25,0}}{\rm{.8)}}\)

Cumulative Density Function cdf of binomial random variable\({\rm{X}}\)with parameters\({\rm{n}}\)and\({\rm{p}}\)is –

\({\rm{B(x;n,p) = P(X}} \le {\rm{x) = }}\sum\limits_{{\rm{y = 0}}}^{\rm{x}} {\rm{b}} {\rm{(y;n,p),}}\;\;\;{\rm{x = 0,1, \ldots ,n}}\)

The theorem is –

\({\rm{b(x;n,p) = }}\left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{{\rm{(1 - p)}}}^{{\rm{n - x}}}}}&{{\rm{,x = 0,1,2, \ldots ,n}}}\\{\rm{0}}&{{\rm{, otherwise }}}\end{array}} \right.\)

Now compute the requested probability as –

\(\begin{aligned}P(7 \leqslant X \leqslant 18) &= B(18;25,0.8) - B(6;25,0.8) \\ &= 0.220 - 0.000 \\ &= 0.220 \\ \end{aligned}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.220}}\).

(c)

The claim is rejected if –

\(\begin{array}{c}{\rm{X}} \le {\rm{7}}\\{\rm{X}} \ge {\rm{18}}\end{array}\)

It is actually correct, so we use probability \({\rm{p = 0}}{\rm{.5}}\) to compute the probability of event –

\({\rm{Rc = }}\){rejecting the claim}

Thus, it is obtained –

\(\begin{aligned} P(Rc) &= P(\{ X \leqslant 7\} \cap \{ X \geqslant 18\} ) = P(X \leqslant 7) + P(X \geqslant 18) \\ &= P(X \leqslant 7) + [1 - P(X \leqslant 17)] \\ &= B(7;25,0.5) + 1 - B(17;25,0.5) \\ &= 0.022 + 1 - 0.978 \\ &= 0.044 \\ \end{aligned} \)

Therefore, the value is obtained as \({\rm{0}}{\rm{.044}}\).

(d)

Complement of event\({\rm{Rc}}\)is event –

\({\rm{8}} \le {\rm{X}} \le {\rm{17}}\)

When\({\rm{p = 0}}{\rm{.6}}\), the probability that the claim is not rejected is –

\(\begin{aligned} P(8 \leqslant X \leqslant 17) &= B(17;25,0.6) - B(7;25,0.6) \\ &= 0.846 - 0.001 \\ &= 0.845 \\ \end{aligned} \)

When\({\rm{p = 0}}{\rm{.8}}\), the probability that the claim is not rejected is –

\(\begin{aligned}P(8 \leqslant X \leqslant 17) &= B(17;25,0.8) - B(7;25,0.8) \\ &= 0.109 - 0.000 \\ & = 0.109 \\ \end{aligned} \)

Therefore, the values are obtained as \({\rm{0}}{\rm{.845}}\) and \({\rm{0}}{\rm{.109}}\).

(e)

From the relation it is obtained –

\({\rm{P(Rc)}} \le {\rm{0}}{\rm{.01}}\)

Find values of random variable\({\rm{X}}\)that would yield such probability.

When the decision rule is –

\(\begin{array}{c}{\rm{X}} \le {\rm{7}}\\{\rm{X}} \ge {\rm{18}}\end{array}\)

Probability of\({\rm{0}}{\rm{.044}}\)is obtained, therefore smaller decision rule is needed, for example –

\(\begin{array}{c}{\rm{X}} \le {\rm{6}}\\{\rm{X}} \ge {\rm{18}}\end{array}\)

Using this, the following probability is obtained –

\(\begin{aligned} P(Rc) &= P(\{ X \leqslant 6\} \cap \{ X \geqslant 18\} ) = P(X \leqslant 6) + P(X \geqslant 18) \\ &= P(X \leqslant 6) + [1 - P(X \leqslant 17)] \\ & = B(6;25,0.5) + 1 - B(17;25,0.5) \\ &= 0.007 + 1 - 0.978 \\ & = 0.029 \\\end{aligned} \)

Which is still not small enough. Then try another decision rule –

\(\begin{array}{c}{\rm{X}} \le {\rm{5}}\\{\rm{X}} \ge {\rm{19}}\end{array}\)

Using this, the following probability is obtained –

\(\begin{aligned} P(Rc) &= P(\{ X \leqslant 5\} \cap \{ X \geqslant 19\} ) = P(X \leqslant 5) + P(X \geqslant 19) \\ &= P(X \leqslant 5) + [1 - P(X \leqslant 18)] \\ &= B(5;25,0.5) + 1 - B(18;25,0.5) \\&= 0.002 + 1 - 0.993 \\ &= 0.009 \\ \end{aligned} \)

Therefore, the last decision rule can be used.